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FBD

John Mallinckrodt wrote:

... Indeed, my reaction (or overreaction, if you will) was primarily
to the original poster's apparent insistence that you *had* to show
the centripetal force on an FBD. ...

This is not different from the free body diagram of an accelerating
Atwood machine block. We show only two forces, W and T; the
m*a is not shown as a vector on the FBD diagram. We find the
net force (W-T) and we equate it to m*a. Right?

If I transferred the m*a to the left side (with the negative sign), and
drew the -m*a vector on FBD, then I would be saying "the particle
is in equilibrium, the sum of all forces is zero". Is this permissible
for an accelerating object?
Ludwik Kowalski